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Pre-Calculus 11: HW 1.4 Geometric Series Sn 
a 1  r n 
1 r
1. Given each geometric sequence, indicate the values of the first term “a”, number of terms “n”, and common
ratio “r”
a) 36  18  9  4.5  2.25  1.125
b) 1  2  4  8  16  32  64  128
a  36
a 1
r 2
n 7
d) 0.125  0.25  0.5  1  2  4  8  16  32
r  0.5
n6
c) 5  10   20   40   80   ......  t11
a  5
r  2
n  11
a  0.125
r 2
5 5
f)   5  ...  40
4 2
5 n 1
40   2 
4
32  2 n 1
a  5/ 4
e) 4   4   4   4   4   4   4   4 
a 4
r  1
2
 2  6  18  ....  486
3
2 n 1
486   3
3
729  3n 1
a  2/3
6  n 1
n 8
g)
r 3
n 7
7n
n 9
r 2
n 6
r  4/3
n 6
5  n 1
6n
27 9
64
h)
  3  ...... 
16 4
9
n 1
64 27  4 
  
9 16  3 
1024  4 
 
243  3 
5  n 1
n 1
a  27 / 16
6n
2. Given each of the following series, find the sum
a) 2.5  5  10  20  40  ....  t8
a  2.5, r  2, n  8
Sn 

a 1  r n 
1 r
2.5 1  28 
1 2
2.5 1  256 

 1
 637.5
b) 8  12  18  27  40.5  .....t9
3
a  8, r  , n  9
2
a 1  r n 
Sn 
1 r
8 1  1.59 

1  1.5
8 1  38.443359375

 0.5
 599.09375
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c) 0.25  0.50  1.0  2.0  4.0 + ... + t10
a  0.25, r  2, n  10
Sn 

a 1  r n 
1 r
0.25 1  210 
1 2
0.25  1023

 1
 255.75
e) 4  8  16  32  64.....  2048
n
a  4, r  2, tn  2048, n  ? S  a 1  r 
n
1 r
tn  a  r n 1
4 1  210 
n 1
2048  4   2 

1 2
512  2n 1
4 1  1024 

9  n 1
 1
10  n
 4092
3
g) 24  12  6  3  ....... 
16
a  24, r  0.5, tn  3 / 16, n  ?
a 1  r n 
Sn 
tn  a  r n 1
1 r
3
n 1
24 1  0.58 
 24   0.5

16
1 2
1
n 1
  0.5
24 1  1024 
128

 1
7  n 1
 23.90625
8n
i) a  3.5, r  0.5, n  10, S10  ?
a  3.5, r  0.5, n  10
Sn 


a 1  r n 
1 r
3.5 1  0.510 
1  0.5
3.5  0.9990234375
0.5
 6.9931640625
2
1 1
1
1

 

+ .... t7
3
3
6
12
24
a  2 / 3, r  0.5, n  7
d)
Sn 
a 1  r n 
1 r
2
7
1   0.5
 3
1   0.5


2
1  0.0078125
 3
1.5
 0.44097222222
f) 3  9  27  81  243  ....  19683
n
a  3, r  3, tn  19683, n  ? S  a 1  r 
n
1 r
tn  a  r n 1
3 1  39 
n 1
19683  3   3

1 3
6561  3n 1
3 1  19683

8  n 1
2
 
9n
 29523
64 32 16
h)

  8  ......  40.5
27 9
3
64
3
a 1  rn 
a
, r  , tn  40.5, n  ? S  
n
27
2
1 r
7
tn  a  r n 1
64 
3 
1



81 64
n 1
27 
2 

 1.5

2 27
1  1.5
n 1
2187  3 
64
 
1  17.0859375
128  2 
27

 0.5
7  n 1
8n
 76.2592592593
j) a  4, r  2, n  6, S6  ?
a  4, r  2, n  6
Sn 

a 1  r n 

1 r
4 1   2 
6

1 2
4  63

 3
 84
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27
2
, r = , n  8, S8  ?
32
3
27
2
a
,r  ,n  8
32
3
a 1  r n 
Sn 
1 r
8
27 
2 
1  
32 
3 

2
1
3
27
 0.96098155768 
32

1
3
 2.4324845679
k) a 
L) a  125, r  0.2, n  7, S7  ?
a  125, r  0.2, n  7
Sn 


a 1  r n 
1 r
125 1  0.27 
1  0.2
125  0.9999872 
0.8
 156.248
3. Given each geometric series, find the value of the missing term:
1
,a  ?
4
a  ?, r  0.25, n  6, S6  341.25
b) S6  567, r 
Sn 
Sn 
a) S6  341.25, r 
a 1  r n 
1 r
a 1  0.256 
341.25 
1  0.25
a  0.99975585937 
341.25 
0.75
256  a
c) The sum of the first 8 terms of a geometric series is
1020 with r  2 . Determine the first term.
a  ?, r  2, n  8, S8  1020
Sn 
a 1  r n 
1020 
1020 

1   2 
a  255 
a 1  r n 
567 
1 r
a 1  0.56 
1  0.5
a  0.984375
567 
0.5
288  a
d) S5  100, S4  87, t5  ?
S5  100, S 4  87, t5  ?
S5  t1  t2  t3  t4  t5
S 4  t1  t2  t3  t4
1 r
a 1   2 
1
,a  ?
2
a  ?, r  0.5, n  6, S6  567
8

S5  S 4  t 4
100  87  t4
13  t4
3
12  a
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4. The sum of the 1st and 2nd term of a geometric sequence is 4 and the sum of the 3rd and 4th term is 36.
Determine the sum of the first 8 terms.
a  ar  4
ar 2  ar 3  36
Factor out any common factors from both equations to simplify it:
a 1  r   4
ar 2 1  r   36
Divide the two equations to solve for the common ratio “r”
ar 2 1  r 
a 1  r 

36
4
ar 2
9
a
r2  9
r  3
There will now be two different possible answers since the common ratio has two different values (+) and (-).
Now solve for “a” for each common ratio:
a 1  r   4
a 1  r   4
a 1   3   4
a 1   3    4
a  2   4
a 4  4
a  2
a 1
Since we have two different “a” values, that means there will be two different sequences and two different sums:
First Sum (a=-2)
S8 
S8 
S8 
For the second sum (a=1)
a 1  r n 
S8 
1 r

2 1    3 
1   3 
8

2 1  6561
S8  3280
4
S8 
S8 
a 1  r n 

1 r
1 1   3
1   3
8

1 1  6561
2
S8  3280
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5. Challenge: In a geometric series, S7  381 and S6  189 , what is the value of the common ratio? Please
show all your work:
There are two separate solutions. In my opinion, the first one is better.
Solution 1:
Since S6  189
189 
a  r 6  1
S7  381
and
r 1
381 
a  r 7  1
r 1
, from here there are usually a few
options. You can either, “add”, “subtract”, or “divide” the two equations to eliminate some variables. So, I decided
to divide them:
S7 381

S6 189

7
381 a  r  1
r 1
Cancel and simplify anything possible:


189
r 1
a  r 6  1

7
127  r  1

63  r 6  1
after we simplify, we know that: 27  1  127
and
26  1  63
Therefore, we conclude that r  2
2nd Method: (not as good)
One way to start this question off is to write out the series:
S7  a  ar  ar 2  ar 3  ar 4  ar 5  ar 6  381
S6  a  ar  ar 2  ar 3  ar 4  ar 5  192
Both series can factor out the constant “a”. Both sums have a greatest common factor of “3”
a 1  r  r 2  r 3  r 4  r 5  r 6   3 127 
a 1  r  r 2  r 3  r 4  r 5   3  63
Then lets assume that a=3
If we subtract both series then:
a  ar  ar 2  ar 3  ar 4  ar 5  ar 6
  a  ar  ar 2  ar 3  ar 4  ar 5 

381
 189 
then substitute a=3
ar 6  192
3r 6  3  64 
r 6  64
r  2
The problem with this solution is that we assume “a” is equal to 3. With 2 variables and only one equation, it
is possible to have more than one solution.
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